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When is convergence in measure useful?

So, as a statistician, the typical use cases I've seen are, in order of "strength" as convergence and convergence in probability to a constant. Intuitively, I consider these to be the ...
Guillaume Dehaene's user avatar
5 votes

This expected value has a minimum!

Another way to do this arises from stochastic dominance. Given a positive random variable $X$ whose p.d.f is bounded above by $1$, create a uniform random variable $U[0,1]$ on the same probability ...
Sarvesh Ravichandran Iyer's user avatar
8 votes

This expected value has a minimum!

Here is a variant of @ConnFus's solution. Let $f(x)$ denote the density function of $X$, and let $g(x) = \mathbf{1}_{[0, 1]}(x)$. Then we note: $\int_{0}^{\infty} [f(x) - g(x)] \, \mathrm{d}x = 0$, $...
Sangchul Lee's user avatar
10 votes

This expected value has a minimum!

Because $X$ is positive, $$ E[X] = \int_0^\infty a(x) \,dx \ge \int_0^1 a(x) \,dx $$ where $a(x) = 1-F(x) = 1 - \int_0^x f(u) \,du$, with $a(0)=1$. Now, $f(x) \le 1 \implies a'(x) \ge -1 \implies a(x)...
leonbloy's user avatar
  • 64.4k
1 vote

How do we know the dual pairing between Lp spaces is well defined?

As long as $q$ is supposed to be the conjugate of $p$ (i.e. $\frac1p + \frac1q = 1$) and $p \geq 1$ then the integral of the product is well-defined thanks to Hölder's inequality, which holds for all ...
Bruno B's user avatar
  • 5,849
1 vote

Bounding $\Vert f\Vert \Vert g\Vert$ by $\Vert wf \Vert^2 +\Vert w^{-1} g\Vert$

The inequality does not hold for any nonconstant $w(x).$ Assume $f$ does not vanish a.e.. For $d=1$ the LHS is equal $$\int\limits_0^1[w^2(x)|f(x)|^2+w^{-2}(x)|g(x)|^2]\,dx\ge 2\int\limits_0^1 |f(x)|\,...
Ryszard Szwarc's user avatar
2 votes
Accepted

Bounding $\Vert f\Vert \Vert g\Vert$ by $\Vert wf \Vert^2 +\Vert w^{-1} g\Vert$

For $f = \frac{1}{w}, g = w$ the left hand side is just $2$, so we only have to pick $w$ such that the right hand side is larger than that. For example when $d = 1$ for any $n \in \mathbb{N}$ we can ...
user23571113's user avatar
  • 1,458
1 vote
Accepted

$\phi$ is continuous and odd, $f$ is Lebesgue integrable, we define $T_nf = \int_{R} \phi(nx)f(x)$. Show that: $lim_{n \to \infty} T_nf = 0$

This can be solved by direct application of Fejer's lemma: Theorem (Féjer): Suppose $\phi$ is a bounded measurable $T$ periodic function ($T>0$). For any $f\in\mathcal{L}_1(\mathbb{R})$ and ...
Mittens's user avatar
  • 40.7k
2 votes

Evaluate $\lim_{n \rightarrow \infty} \int_0^{\infty} ne^{-nx} \frac{x^2+1}{x^2+x+1}dx$

For this integral a change of variables allows write the integral of interest as integrals of a rescaling of the function $p(x)=\frac{x^1+1}{x^2+x+1}$ with respect to a fixed probability distribution....
Mittens's user avatar
  • 40.7k
2 votes

Evaluate $\lim_{n \rightarrow \infty} \int_0^{\infty} ne^{-nx} \frac{x^2+1}{x^2+x+1}dx$

Too long for a comment and just for your curiosity. This looks fine to me. If you want to go further with bounds, write $$\frac{x^2+1}{x^2+x+1}=\frac{x^2+1}{(x-a)(x-b)}=1+\frac{a^2+1}{(a-b) (x-a)}-\...
Claude Leibovici's user avatar
2 votes

Show $0<\epsilon<1$ there exists some $\delta_{p,\epsilon}>0$ such that $m(\{x\in X:|f(x)|>\epsilon\})\geq\delta_{p,\epsilon}$ for each $f\in E_p$.

Notice that for $f\in E_p$ and $0<\varepsilon<1$, \begin{align}1&=\int_X|f|=\int_{\{|f|\leq\varepsilon\}}|f|+\int_{\{|f|>\varepsilon\}}|f|\\ &\leq \varepsilon +\|f\|_p\big(\mu(|f|>\...
Mittens's user avatar
  • 40.7k
1 vote

Minkowski inequality of infinite sum

Not as nice as the other answers as it requires sigma-finiteness, but here is a possibility which explores some beautiful results in measure theory! If one has that $X$ is a $\sigma$-finite space one ...
Kadmos's user avatar
  • 2,234
0 votes

Riemann-Lebesgue lemma

We had the following lemma in class: [Riemann-Lebesgue Lemma] Let $h \in \mathcal{C}(\mathbb{R})$ be a $T$-periodic function, with mean value $m = \frac{1}{T} \int_{0}^{T} h(x) \, dx.$ Then for all $g ...
arridadiyaat's user avatar
0 votes

Riemann integral counterexample to dominated convergence theorem?

What if we add the premise that $f$ is also Riemann integrable? If, in addition to that, the interval $I$ is closed and bounded and all $f_n$s are bounded, Arzelà's dominated convergence theorem ...
Nil Admirari's user avatar

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